Question

Question #05115

Answer 1

`"1400 mL"`

Explanation:

First thing first, you're actually dealing with sodium lactate, not sodium "locate".

Sodium lactate, `"NaC"_3"H"_5"O"_3`, is the salt of lactic acid, `"C"_3"H"_6"O"_3`, and completely dissociates in aqueous solution to produce the lactate anions, `"C"_3"H"_5"O"_3^(-)`, the conjugate base of lactic acid.

This means that you're dealing with a buffer solution that consists of lactic acid, a weak acid, and the lactate anion, its conjugate base.

Now, you can use the Henderson - Hasselbalch equation to find the ratio that must exist between the concentration of the weak acid and that of the conjugate base so that the pH of the solution is equal to `4.00`.

`color(blue)("pH" = pK_a = log( (["conjugate base"])/(["weak acid"]))`

Now, in order to get the acid's `pK_a`, you need to know that acid dissociation constant, `K_a`, which for lactic acid is listed as

`K_a = 1.38 * 10^(-4)`

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

The key here is to first find the number of moles of lactic acid you have in the initial solution. To do that, use its molarity and volume

`color(blue)(c = n/V implies n = c * V)`

`n_"lactic acid" = "1.0 M" * "1.0 L" = "1.0 moles"`

Now, the important thing to realize here is that after the sodium lactate solution is added, the volume of both solutions will be the same.

This means that you can go by the nubmer of moles of weak acid and of conjugate base needed to make that buffer. Assuming that you need to add `x` liters of sodium lactate solution, you will get

`V_"buffer" = V_"lactic acid" + V_"sodium lactate"`

`V_"buffer" = "1.0 L" + xcolor(white)(x) "L" = (1.0 + x)"L"`

This means that the Henderson - Hasselbalch equation wil take the form

`"pH" = -log(K_a) + log ( ("moles of conjugate base"/color(red)(cancel(color(black)((1.0 + x)))))/("moles of weak acid"/color(red)(cancel(color(black)((1.0 + x))))))`

`4.00 = 3.86 + log("moles of conjugate base"/"moles of weak acid")`

This is equivalent to

`log("moles of conjugate base"/"moles of weak acid") = 0.14`

To get rid of the log, use

`"moles of conjugate base"/"moles of weak acid" = 10^(0.14)`

`"moles of conjugate base"/"1.0 moles" = 1.38`

This means that you need to add

`n_"lactate" = "1.00 moles" * 1.38 = "1.38 moles"`

So, what volume of the sodium lactate solution would be needed to deliver this many moles of conjguate base to the buffer?

`c = n/V implies V = n/c`

`V_"lactate" = (1.38 color(red)(cancel(color(black)("moles"))))/(1.0color(red)(cancel(color(black)("moles")))/"L") = "1.38 L"`

Do a quick check to make sure that the concentrations of the two species would still produce a pH equal to `4`

`["C"_3"H"_6"O"_3] = "1.0 moles"/((1.0 + 1.38)"L") = "0.4202 M"`

`["C"_3"H"_5"O"_3^(-)] = "1.38 moles"/"2.38 L" = "0.5798 M"`

The pH of the buffer will be

`"pH" = 3.86 + log( (["C"_3"H"_5"O"_3^(-)])/(["C"_3"H"_6"O"_3]))`

`"pH" = 3.86 + log( (0.5798color(red)(cancel(color(black)("M"))))/(0.4202color(red)(cancel(color(black)("M")))))`

`"pH" = 3.9998 ~~ 4`

Therefore, the volume of the sodium lactate solution that must be dded to the lactic acid solution is equal to `"1.38 L"`. Expressed in mililiters, this will be equal to

`V_"lactate" = "1400 mL" ->` rounded to two sig figs